Question

When we jump on a heap of sand, we don't get hurt but we get hurt when the floor is concrete. Explain.

Answer 1

Hint: To answer this question, use the formula for impulse on a body in terms of force and change in time period. Then, use the formula for impulse in terms of changing momentum. Combining the above two formulas, we can get the required relation between force on the body, change in momentum and time period. Use this relation, to answer the given question.
Formula used:
$Impulse=Force \times time$
$J= m \Delta v$

Complete answer:
We know, Impulse is given by,
$Impulse=Force \times time$
$\therefore J= F \times \Delta t$
Rearranging the above equation we get,
$F= \dfrac {J}{\Delta t}$ …(1)
Also, impulse is related to change in momentum which is given by,
$J= m \Delta v$ …(2)
Thus, equation. (1) turns to,
$F= \dfrac {m \Delta v}{\Delta t}$ ,,,(3)
When we jump on a heap of sand, there is a small change in momentum which causes a small impulse. And the time taken by the body to come to rest is larger. Thus, applying the above equation in this case, we can infer that the force exerted on the body is small and we don’t get hurt.
But, when we jump on the floor which is made up of concrete, the rate of change of momentum is very high and thus, the feet of the person comes to rest instantaneously. Applying these conditions to the equation. (3), it can be inferred that the force exerted on the body is large. Thus, when we jump on the floor made up of concrete, we get hurt.
Therefore, when we jump on a heap of sand, we don’t get hurt but we get hurt when the floor is concrete.

Note:
Students may get confused between jerk and impulse. They must remember that the impulse on a body is given by the change in momentum while jerk is given by the change in acceleration. While, considering impulse, mass of the body is taken into account but while considering jerk, mass does not come in the picture. Mathematical expression for jerk is given by,
$J=\dfrac { \partial \vec { a } }{ \partial t }$
$\Rightarrow J=\dfrac { { \partial }^{ 2 }\vec { v } }{ \partial t }$
$\Rightarrow J=\dfrac { { \partial }^{ 3 }\vec { s } }{ \partial t }$